If $\angle CMP=90^{\circ}$, find $x$. 
Let $ABC$ an equilateral triangle, $M \in [CA]$, $N\in [AB]$, $P\in [BC]$ s.t $\angle CBM=x, \angle AMN=2x$ and angle $ \angle BNP=3x$. If $\angle CMP=90^{\circ}$, find $x$.

I tried a trigonometric solution as below but I am stuck. I noticed that $\triangle BNM$ is isosceles and I think that $x=15^{\circ}.$

 A: Here is a possible solution of the problem, showing the uniqueness of $x$ between $0^\circ$ and $30^\circ$ with the given property, then using the OP guess $x=15^\circ$, and showing this guess.

For the uniqueness, consider the points $M,N,P_1,P_2$ defined as in the picture, seen as functions of $x$.

Consider the segments $BC$, $CA$, $AB$ oriented as given by the vectors
$\vec{BC}$,
$\vec{CA}$,
$\vec{AB}$, and extend them to axes of coordinates for the corresponding lines, so that these segments are of (oriented) length one.
Consider the points $M,N,P_1,P_2$ on the sides of $\Delta ABC$, i.e. on these axes as in the picture, to be defined by the following conditions as functions of $x$,

*

*$M=M(x)$ with $\widehat{MBC}=x$,
then
$N=N(x)$ with $\widehat{NMA}=2x$,
then
$P_1=P_1(x)$ with $\widehat{P_1NB}=2x$,
and

*$P_2=P_2(x)$ with $\widehat{P_2MC}=90^\circ$.

Then the functions $M,N,P_1$ are increasing, and $P_2$ decreasing.
In particular, $(P_1-P_2(x)=P_1(x)-P_2(x)$, with minus operation taken from the axis $\vec{BC}$, is an increasing continuous function, and
$(P_1-P_2)(0^\circ)=B-C=0-1=-1$,
$(P_1-P_2)(30^\circ)=C-B=1-0=+1$, so there exists a unique argument $x$ between $0^\circ$ and $30^\circ$ with $P_1(x)=P_2(x)$.

It is easy to show that $x=15^\circ$ verifies $P_1(x)=P_2(x)$.
This is so, since for this $x$ the triangle $\Delta BMN$ is right and isosceles, and $NP_1\perp BM$ becomes the perpendicular bisector of $BM$. Thus $\widehat{P_1BM}=\widehat{P_1MB}=x=15^\circ$, making $\widehat{CP_1M}=2x=30^\circ$, i.e. $P_1M\perp AC$, and thus $P_1=P_2$.
$\square$
